11.3 Lebesgue integration

中文

Let $h$ be a simple function on a measure space $(X,\mathcal{M},\mu )$

$$h=\sum _{i=1}^n{a}_i{\chi }_{A_i}(a_i\in \dot{\mathbb{R}})$$

where $A_i=h^{-1}(a_i)$ are measurable subsets of $X$. We define its integral to be

$$\int h\mathrm{d}\mu =\sum _{i=1}^n{a}_i\mu (A_i).$$

This integral only gives a finite answer if the measure of all sets $A_i$ is finite, and in some cases it may not have a sensible value at all. For example $h:\mathbb{R}\to \mathbb{R}$ defined by

$$h(x)=\{\begin{array}{ll}1& \text{ if }x>0\\ -1& \text{ if }x<0\end{array}$$

has integral $\infty +(-\infty )$ which is not well-defined.

If $h:X\to \mathbb{R}$ is a simple function, then for any constant $b$ we have

$$\int bf\mathrm{d}\mu =b\int f\mathrm{d}\mu .\text{\hspace{1em}(11.8)}$$

If $\begin{array}{r}g={\sum }_{j=1}^m{b}_j{\chi }_{B_j}\end{array}$ is another simple function then $f+g$ is a simple function

$$f+g=\sum _{i=1}^n\sum _{j=1}^m(a_i+b_j){\chi }_{A_i\cap B_j},$$

and has integral

$$\int (f+g)\mathrm{d}\mu =\int f\mathrm{d}\mu +\int g\mathrm{d}\mu .\text{\hspace{1em}(11.9)}$$

It is best to omit any term from the double sum where $a_i+b_j=0$ , else we may face the awkward problem of assigning a value to the product $0\infty$.

Exercise

Prove Eq. (11.8).

For any pair of functions $f,g:X\to \mathbb{R}$ we write $f\le g$ to mean $f(x)\le g(x)$ for all $x\in X$ . If $h$ and $h^{\prime }$ are simple functions such that $h\le h^{\prime }$ then

$$\int h\mathrm{d}\mu \le \int h^{\prime }\mathrm{d}\mu .$$

This follows immediately from the fact that $h-h^{\prime }$ is a simple function that is non-negative everywhere, and therefore has an integral $\ge 0$

Taking the measure on $\mathbb{R}$ to be Lebesgue measure, the integral of a non-negative measurable function $f:X\to \mathbb{R}$ is defined as

$$\int f\mathrm{d}\mu =\sup \int h\mathrm{d}\mu$$

Figure 11.3 Integral of a non-negative measurable function

where the supremum is taken over all non-negative simple functions $h:X\to \mathbb{R}$ such that $h\le f$ (Fig. 11.3) . If $E\subset X$ is a measurable set then $f{\chi }_E$ is a measurable function on $X$ that vanishes outside $E$. We define the integral of $f$ over $E$ to be

$${\int }_{E}f\mathrm{d}\mu =\int f{\chi }_E\mathrm{d}\mu .$$

Exercise

Show that for any pair of non-negative measurable functions $f$ and $g$

$$f\ge g⟹\int fd\mu \ge \int gd\mu .\text{\hspace{1em}(11.10)}$$

The following theorem is often known as the monotone convergence theorem:

Theorem 11.7 · Monotone convergence theorem (Beppo Levi)

(Beppo Levi) If $f_n$ is an increasing sequence ofnon-negative measurable real-valuedfunctions on $X$, $f_{n+1}\ge f_n$ , such that $f_n(x)\to f(x)$ for all $x\in X$ then

$$\underset{n\to \infty }{\lim }\int f_{n}d\mu =\int fd\mu .$$

Proof

From the comments before Theorem 11.2 we know that $f$ is a measurable function, as it is the limit of a sequence of measurable functions. If $f$ has a finite integral then, by definition, for any $ϵ>0$ there exists a simple function $h:X\to \mathbb{R}$ such that $0\le h\le f$ and $\int f\mathrm{d}\mu -\int h\mathrm{d}\mu <ϵ$ . For any real number $0<c<1$ let

$$E_n=\{x\in X|f_n(x)\ge ch(x)\}=(f_n-ch{)}^{-1}([0,\infty )),$$

clearly a measurable set for each positive integer $n$. Furthermore, since $f_n$ is an increasing sequence of functions we have

$$E_n\subset E_{n+1}\text{ and }X=\bigcup _{n=1}^{\infty }{E}_n,$$

since every point $x\in X$ lies in some $E_n$ for $n$ big enough. Hence

$$\int f\mathrm{d}\mu \ge \int f_n\mathrm{d}\mu \ge c\int h{\chi }_{E_n}\mathrm{d}\mu .$$

If $\begin{array}{r}h={\sum }_i{a}_i{\chi }_{A_i}\end{array}$ then

$$\int h{\chi }_{E_n}\mathrm{d}\mu =\int \sum _i{a}_i{\chi }_{A_i\cap E_n}\mathrm{d}\mu =\sum _i{a}_i\mu (A_i\cap E_n).$$

Hence, by Theorem 11.3,

$$\underset{n\to \infty }{\lim }\int h{\chi }_{E_n}\mathrm{d}\mu =\sum _i{a}_i\mu (A_i\cap X)=\sum _i{a}_i\mu (A_i)=\int h\mathrm{d}\mu ,$$

so that

$$\int f\mathrm{d}\mu \ge \underset{n\to \infty }{\lim }\int f_n\mathrm{d}\mu \ge c\int h\mathrm{d}\mu \ge c\int f\mathrm{d}\mu -cϵ.$$

Since $c$ can be chosen arbitrarily close to 1 and $ϵ$ arbitrarily close to $0,$ , we have

$$\int f\mathrm{d}\mu \ge \underset{n\to \infty }{\lim }\int f_n\mathrm{d}\mu \ge \int f\mathrm{d}\mu ,$$

which proves the required result.

Exercise

How does this proof change if $\int f\mathrm{d}\mu =\infty ?$

Using the result from Theorem 11.2, that every positive measurable function is a limit of increasing simple functions, it follows from Theorem 11.7 that simple functions can be replaced by arbitrary measurable functions in Eqs. Eq. (11.8) and Eq. (11.9).

Theorem 11.8 · Vanishing integral of a nonnegative measurable function

The integral of a non-negative measurable function $f\ge 0$ vanishes if and only if $f(x)=0$ almost everywhere.

Proof

If $f(x)=0a.e.$ ., let $\begin{array}{r}h={\sum }_i{a}_i{\chi }_{A_i}\ge 0\end{array}$ be a simple function such that $h\le f$ Every set $A_i$ must have measure zero, else $f(x)>0$ on a set of positive measure. Hence $\int f\mathrm{d}\mu =\sup \int h\mathrm{d}\mu =0.$

Conversely, suppose $\int f\mathrm{d}\mu =0$ . Let $E_n=\{x\mid f(x)\ge 1/n\}$ . These are an increasing sequence of measurable sets, $E_{n+1}\supset E_n$ , and

$$f\ge \frac{1}{n}{\chi }_{A_n}.$$

Hence

$$\int \frac{1}{n}{\chi }_{A_n}\mathrm{d}\mu =\frac{1}{n}\mu (A_n)\le \int f\mathrm{d}\mu =0,$$

which is only possible if $\mu (A_n)=0$ . By Theorem 11.3 it follows that

$$\mu (\{x\mid f(x)>0\})=\mu \left(\bigcup _{n=1}^{\infty }{A}_n\right)=\underset{n\to \infty }{\lim }\mu \left(A_n\right)=0.$$

Hence $f(x)=0$ almost everywhere.

Integration may be extended to real-valued functions that take on positive or negative values. We say a measurable function $f$ is integrable with respect to the measure $\mu$ if both its positive and negative parts, $f^{+}$ and $f^{-}$ , are integrable. The integral of $f$ is then defined as

$$\int f\mathrm{d}\mu =\int f^{+}\mathrm{d}\mu -\int f^{-}\mathrm{d}\mu .$$

If $f$ is integrable then so is its modulus $|f|=f^{+}+f^{-}$ , and

$$\left|\int f\mathrm{d}\mu \right|\le \left|\int f^{+}\mathrm{d}\mu -\int f^{-}\mathrm{d}\mu \right|\le \left|\int f^{+}\mathrm{d}\mu +\int f^{-}\mathrm{d}\mu \right|\le \int |f|\mathrm{d}\mu .\text{\hspace{1em}(11.11)}$$

Hence a measurable function $f$ is integrable if and only if $|f|$ is integrable.

A function $f:\mathbb{R}\to \mathbb{R}$ is said to be Lebesgue integrable if it is measurable and integrable with respect to the Lebesgue measure on $\mathbb{R}$. As for Riemann integration it is common to use the notations

$$\int f(x)\mathrm{d}x={\int }_{-\infty }^{\infty }f(x)\mathrm{d}x\equiv \int f\mathrm{d}\mu ,$$

and for integration over an interval $I=[a,b]$

$${\int }_a^{b}f(x)\mathrm{d}x\equiv {\int }_{I}f\mathrm{d}\mu .$$

Riemann integrable functions on an interval $[a,b]$ are Lebesgue integrable on that interval. A function $f$ is Riemann integrable if for any $ϵ>0$ there exist step functions $h_1$ and $h_2-$ simple functions that are constant on intervals – such that $h_1\le f\le h_2$ and

$${\int }_a^b{h}_2(x)\mathrm{d}x-{\int }_a^b{h}_1(x)\mathrm{d}x<ϵ.$$

By taking $H_n=\sup (h_{i1})$ for the sequence of functions $(h_{i1},h_{i2})(i=1,\dots ,n)$ defined by $ϵ=1,\frac{1}{2},\dots ,\frac{1}{n}.$ , it is straightforward to show that the $H_n$ are simple functions the supremum of whose integrals is the Riemann integral of $f.$ Hence $f$ is Lebesgue integrable, and its Lebesgue integral is equal to its Riemann integral. The difference between the two concepts of integration is that for Lebesgue integration the simple functions used to approximate a function $f$ need not be step functions, but can be constant on arbitrary measurable sets. For example, the function on [0, 1] defined by

$$f(x)=\{\begin{array}{ll}1& \text{ if }x\text{ is irrational }\\ 0& \text{ if }x\text{ is rational }\end{array}$$

is certainly Lebesgue integrable, and since $f=1a.e$ . its Lebesgue integral is 1. It cannot, however, be approximated in the required way by step functions, and is not Riemann integrable.

Exercise

Prove the last statement.

Theorem 11.9 · Linearity of the Lebesgue integral

If $f$ and $g$ are Lebesgue integrable real functions, then for any $a$, $b\in \mathbb{R}$ the function $af+bg$ is Lebesgue integrable and for any measurable set $E$

$${\int }_E(af+bg)\mathrm{d}\mu =a{\int }_{E}f\mathrm{d}\mu +b{\int }_{E}g\mathrm{d}\mu .$$

The proof is straightforward and is left as an exercise (see problems at end of chapter).

Lebesgue’s dominated convergence theorem

One of the most important results of Lebesgue integration is that, under certain general circumstances, the limit of a sequence of integrable functions is integrable. First we need a lemma, relating to the concept oflim sup of a sequence of functions, defined in the paragraph prior to Theorem 11.2.

Lemma 11.10 · Fatou’s lemma

(Fatou) If $(f_n)$ is any sequence ofnon-negative measurable functions de fined on the measure space $(X,\mathcal{M},\mu )$ , then

$$\int \underset{n\to \infty }{\mathrm{liminf}}{f}_n\mathrm{d}\mu \le \underset{n\to \infty }{\mathrm{liminf}}\int f_n\mathrm{d}\mu .$$

Proof

The functions $\begin{array}{r}{G}_n(x)={\inf }_{k\ge n}{f}_k(x)\end{array}$ form an increasing sequence of non-negative measurable functions such that $G_n\le f_n$ for all $n$. Hence the lim inf of the sequence $(f_n)$ is the limit of the sequence $G_n$

$$\underset{n\to \infty }{\mathrm{liminf}}{f}_n=\underset{n}{\sup }{G}_n=\underset{n\to \infty }{\lim }{G}_n.$$

By the monotone convergence theorem 11.7,

$$\underset{n\to \infty }{\lim }\int G_n\mathrm{d}\mu =\int \underset{n\to \infty }{\mathrm{liminf}}{f}_n\mathrm{d}\mu$$

while the inequality $G_n\le f_n$ implies that

$$\int G_n\mathrm{d}\mu \le \int f_n\mathrm{d}\mu .$$

Hence

$$\int \underset{n\to \infty }{\mathrm{liminf}}{f}_n\mathrm{d}\mu =\underset{n\to \infty }{\lim }\int G_n\mathrm{d}\mu \le \underset{n\to \infty }{\mathrm{liminf}}\int f_n\mathrm{d}\mu .$$

Theorem 11.11 · Lebesgue dominated convergence theorem

(Lebesgue) Let $(f_n)$ be any sequence of real-valuedmeasurable functions defined on the measure space $(X,\mathcal{M},\mu )$ that converges almost everywhere to a function $f$. If there exists a positive integrable function $g:X\to \mathbb{R}$ such that $|f_n|<g$ for all $n$ then

$$\underset{n\to \infty }{\lim }\int f_{n}d\mu =\int fd\mu .$$

Proof

The function $f$ is measurable since it is the limit a.e. of a sequence of measurable functions, and as $|f_n|<g$ all functions $f_n$ and $f$ are integrable with respect to the measure $\mu$. Apply Fatou’s lemma 11.10 to the sequence of positive measurable functions $g_n=$ $2g-|f_n-f|>0$

$$\int \underset{n\to \infty }{\mathrm{liminf}}(2g-|f_n-f|)\mathrm{d}\mu \le \underset{n\to \infty }{\mathrm{liminf}}\int (2g-|f_n-f|)\mathrm{d}\mu .$$

Since $\int g\mathrm{d}\mu <\infty$ and lim inf $|f_n-f|=\lim |f_n-f|=0$ , we have

$$0\le \underset{n\to \infty }{\mathrm{liminf}}\int -|f_n-f|\mathrm{d}\mu =-\underset{n\to \infty }{\mathrm{limsup}}\int |f_n-f|\mathrm{d}\mu .$$

Since $|f_n-f|>0$ this is only possible if

$$\underset{n\to \infty }{\lim }\int |f_n-f|d\mu =0.$$

Hence

$$\left|\int (f_n-f)\mathrm{d}\mu \right|\le \int |f_n-f|\mathrm{d}\mu \to 0,$$

so that $\int f_n\mathrm{d}\mu \to \int f\mathrm{d}\mu$ , as required.

The convergence in this theorem is said to be dominated convergence, $g$ being the dominating function. An attractive feature of Lebesgue integration is that an integral over an unbounded set is defined exactly as for a bounded set. The same is true of unbounded integrands. This contrasts sharply with Riemann integration where such integrals are not defined directly, but must be defined as ‘improper integrals’ that are limits of bounded functions over a succession of bounded intervals. The concept of an improper integral is not needed at all in Lebesgue theory. However, Lebesgue’s dominated convergence theorem can be used to evaluate such integrals as limits of finite integrands over finite regions.

Example 11.8

The importance of a dominating function is shown by the following example. The sequence of functions $\left({\chi }_{[n,n+1]}\right)$ consists of a ‘unit hump’ drifting steadily to the right and clearly has the limit $f(x)=0$ everywhere. However it has no dominating function and the integrals do not converge

$$\int {\chi }_{[n,n+1]}\mathrm{d}\mu ={\int }_n^{n+1}1\mathrm{d}x=1\nrightarrow \int 0\mathrm{d}x=0.$$

We mention, without proof, the following theorem relating Lebesgue integration on higher dimensional Euclidean spaces to multiple integration.

Theorem 11.12 · Fubini’s theorem

(Fubini) If $f:{\mathbb{R}}^2\to \mathbb{R}$ is a Lebesgue measurable function, then for each $x\in \mathbb{R}$ the function $f_x(y)=f(x,y)$ is measurable. Similarly for each $y\in \mathbb{R}$ the function $f_y^{\prime }(x)=f(x,y)$ is measurable on $\mathbb{R}$. It is common to write

$$\int f(x,y)\mathrm{d}y\equiv \int f_x\mathrm{d}\mu and\int f(x,y)\mathrm{d}x\equiv \int f_y^{\prime }\mathrm{d}\mu .$$

Then

$$\begin{array}{rl}\int \int f(x,y)\mathrm{d}x\mathrm{d}y& \equiv \int f\mathrm{d}{\mu }^2=\int \left(\int f(x,y)\mathrm{d}y\right)\mathrm{d}x\\ & =\int \left(\int f(x,y)\mathrm{d}x\right)\mathrm{d}y.\end{array}$$

The result generalizes to a product of an arbitrary pair of measure spaces. For a proof see, for example, [1, 2].

Problems

Problem 11.12

Show that if $f$ and $g$ are Lebesgue integrable on $E\subset \mathbb{R}$ and $f\ge ga.e$ ., then

$${\int }_{E}f\mathrm{d}\mu \ge {\int }_{E}g\mathrm{d}\mu .$$

Problem 11.13

Prove Theorem 11.9.

Problem 11.14

If $f$ is a Lebesgue integrable function on $E\subset \mathbb{R}$ then show that the function $\psi$ defined by

$$\psi (a)=\mu (\{x\in E||f(x)|>a\})=O(a^{-1})asa\to \infty .$$